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anonymous
 4 years ago
Find all distinct roots(real or complex) of
z^2 + (9+3i)z + (24+11i) in the form a+bi.
Hint: You may have to complete the square
anonymous
 4 years ago
Find all distinct roots(real or complex) of z^2 + (9+3i)z + (24+11i) in the form a+bi. Hint: You may have to complete the square

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://answers.yahoo.com/question/index?qid=20121124223053AAchfPc the bottom explanation of this link might be it, but i dont understand it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0have you figured anything out?

sirm3d
 4 years ago
Best ResponseYou've already chosen the best response.0\[\large z^2+(9+3i)z+(2411i)=0\]\[\large z^2+(9+3i)z=(2411i)\]to complete squares, add (b/2)^2 to both sides, where b=9+3i \[\large z^2+(9+3i)z+\left( \frac{ 9+3i }{ 2 } \right)^2=(2411i)+\left( \frac{ 9+3i }{ 2 } \right)^2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah im at this step... what do you do next?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(z + (\frac{ 9 + 3i }{ 2 }))^2 = 24 + 11i + 18 +\frac{ 27i }{ 2 }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[z + (\frac{ 9+3i }{ 2 }) = \pm \sqrt{\frac{ 12 + 5i}{ 2 }} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i dont get how the person completed the square in the yahoo link..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[z = (\frac{ 9+3i }{ 2}) \pm \sqrt{\frac{ 12 + 5i }{ 2 }}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it would be so much easier if the square root wasnt there...we could just add them up easily

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0from the yahoo answers link: z² + (2−4i)z = (11+10i) z² + (2−4i)z + (3 4i) = (11+10i) + (3 4i)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what i dont get is where the (34i) comes from
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